A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compound?
1 Answer
Explanation:
!! SHORT ANSWER !!
Grab a periodic table and look at the molar masses of the two elements. You have
#"For Ca: " "40.078 g mol"^(-1)#
#"For Cl: " "35.453 g mol"^(-1)#
Notice that the two molar masses are relatively close to each other. Now, the percent composition of the compound tells you that you get about one part calcium per two parts chlorine by mass.
You can thus say that for every mole of this compound, you will get one mole of calcium and two moles of chlorine. Therefore, the empirical formula for this compound will be
#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#
!! LONG ANSWER !!
Here's how you can double-check your result. Let's assume that you have a
#"36.1 g"# of calcium,#"Ca"#
#"63.9 g"# of chlorine,#"Cl"#
Use the molar masses of the two elements to figure out how many moles of each you have in this sample
#"For Ca: " 36.1 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.901 moles Ca"#
#"For Cl: " 63.9 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "1.802 moles Cl"#
To get the mole ratio that exists between the two elements in the compound, divide both values by the smallest one
#"For Ca: " (0.901 color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 1#
#"For Cl: " (1.802color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 2#
Once again, the empirical formula of the compound, which tells you the smallest whole number ratio that exists between the elements that make up the compound, comes out to be
#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#
